An interesting Math problem

Sat, 22 Mar 2025 00:00:00 +0000

📘 Solve the equation:

\[ \sqrt{5 - x} = 5 - x^2 \]

Square both sides:

\[ (\sqrt{5 - x})^2 = (5 - x^2)^2 \] \[ 5 - x = 25 + x^4 - 10x^2 \]

Move all terms to one side:

\[ 0 = x^4 - 5 - 10x^2 + x + 25 \]

We write this as a quadratic in terms of \(5\):
Let:

Use the quadratic formula:

\[ 5 = \frac{(2x^2 + 1) \pm \sqrt{(2x^2 + 1)^2 - 4(x^4 + x)}}{2} \]

We are solving:

\[ 5 = \frac{2x^2 + 1 \pm \sqrt{(-2x^2 - 1)^2 - 4(x^4 + x)}}{2} \]

Now simplify the discriminant:

\[ = \frac{2x^2 + 1 \pm \sqrt{4x^2 - 4x + 1}}{2} \]

Factor the square root:

\[ = \frac{2x^2 + 1 \pm (2x - 1)}{2} \]

Now evaluate the two cases from here, which can be solve using standard quard formula.

Using the quadratic formula:

\[ x^2 + x - 5 = 0 \Rightarrow x = \frac{-1 \pm \sqrt{1^2 + 4 \cdot 5}}{2} = \frac{-1 \pm \sqrt{21}}{2} \]

\[ x^2 - x - 4 = 0 \Rightarrow x = \frac{1 \pm \sqrt{1^2 + 4 \cdot 4}}{2} = \frac{1 \pm \sqrt{17}}{2} \]

Check all four roots against the original equation. Only two satisfy it:

\[ \boxed{\frac{1}{2}(1 - \sqrt{17}) \quad \text{and} \quad \frac{1}{2}(\sqrt{21} - 1)} \]